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2 years ago

-4 + -7 (-2) =? Minus equations always give me trouble :/

Mathematics

2 answers:

USPshnik [31]2 years ago

8 0

<span>-4 + -7 (-2)
= -4 + 14 (a negative multiplies a negative equal positive)
= 10</span>

steposvetlana [31]2 years ago

6 0

The answer is 10.
-4+(-7)(-2)=
-4+14=
10

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Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ

solong [7]

Answer:

The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

Step-by-step explanation:

We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

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2 years ago

Circle C has radius of 10 cm. Each of points B and D is on the midpoint of the radius. Find the area of the shaded region.

olya-2409 [2.1K]

Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(

7 0

2 years ago

Read 2 more answers

At a carnival, a particular game requires the player to spin a wheel. When a child plays, the game operator allows them to conti

Bumek [7]

Answer:

(a) and (e) are not valid.

Step-by-step explanation:

a) The number of traials is not fixed. In fact, the random variable counts the total number of trials.

b) This is true, the result of each trial is independent of the others.

c) This is also true. The outcomes are failure and success.

d) This is true as well, all trials are identically distributed.

e) Since (a) is not valid, this cant be true.

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2 years ago

The rule below shows how Mrs. Rousseau's long-distance phone company computes her monthly bill. Which answer expresses that rule

Ludmilka [50]

So then n=number of long distance calls, then add the flat rate

long distance cost=number of calls times cost per clll or n times 0.42 or 0.42n
the flat rate is 24

so total is c so
c=24+0.42n

4 0

2 years ago

In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 time

ziro4ka [17]

Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

__________________________________________________

First shot Made Second shot missed Total

__________________________________________________

Made 152 33 185

Missed 37 8 45

__________________________________________________

Total 189 41 230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot Made Second shot missed

_____________________________________________________

Made 189(185)/230=152.0217 41(185)/230=32.9783

Missed 189(45)/230=36.9783 41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

$\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}$

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

$\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881$

Now with the calculated value we can find the degrees of freedom

$df=(r-1)(c-1)=(2-1)(2-1)=1$ on this case r means the number of rows and c the number of columns.

Now we can calculate the p value

$p_v =P(\chi^2 >0.00008881)=0.9925$

On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.

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2 years ago